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Lie antialgebras: premices
Valentin Ovsienko
To cite this version:
Lie antialgebras:
pr´emices
V. Ovsienko
Abstract
The main purpose of this work is to develop the basic notions of the Lie theory for commutative algebras. We introduce a class of Z2-graded commutative but not
associative algebras that we call “Lie antialgebras”. These algebras form a very special class of Jordan superalgebras, they are closely related to Lie (super)algebras and, in some sense, link together commutative and Lie algebras. The main notions we define in this paper are: representations of Lie antialgebras, an analog of the Lie-Poisson bivector (which is not Poisson) and central extensions. We will explain the geometric origins of Lie antialgebras and provide a number of examples. We also classify simple finite-dimensional Lie antialgebras. This paper is a new version of the unpublished preprint [10].
Contents
1 Introduction 2
1.1 The definition . . . 2
1.2 Examples . . . 3
1.3 The main properties . . . 5
2 Lie antialgebras and symplectic geometry 6 2.1 The algebra K3 and the odd bivector Λ . . . 6
2.2 The full derivation algebra AK(1) . . . 8
2.3 Representation of AK(1) by tangent vector fields . . . 13
2.4 A pair of symplectic forms on R4|2 and the algebras K 3(C) and AK(1)C . . 15
3 Elements of the general theory 18 3.1 Relation to Lie superalgebras . . . 18
3.2 Representations and modules . . . 20
3.3 The odd Lie-Poisson type bivector . . . 23
3.4 Central extensions . . . 24
4 Classification results 28 4.1 Classification of finite-dimensional simple Lie antialgebras . . . 28
1
Introduction
Let (a, ·) be a commutative Z2-graded algebra over K = R or C, that is, a = a0⊕ a1, such
that ai· aj ⊂ ai+j and for all homogeneous elements x, y one has
x · y = (−1)p(x)p(y)y · x, (1.1) where p is the Z2-valued parity function p|ai = i. In particular, the space a0 ⊂ a is a
commutative subalgebra while the bilinear map a1× a1 → a0 is skew-symmetric.
Typical examples of Z2-graded commutative algebras are the algebras of differential
forms on manifolds, or, more generally, the algebras of functions on supermanifolds; these algebras are of course associative and, in particular, the space a1 is an a0-module.
Exam-ples of commutative but not associative algebras are Jordan algebras.
The algebras considered in this paper are not associative, however the subalgebra a0
will always be associative. In this sense, we suggest an alternative way to extend the associativity condition of a0 to all of the a. We will try to convince the reader this leads
to algebras that have quite remarkable properties.
1.1
The definition
A Z2-graded commutative algebra a is called a Lie antialgebra if it satisfies the following
identities:
α · (β · γ) = (α · β) · γ, (1.2)
α · (β · a) = 12(α · β) · a, (1.3)
α · (a · b) = (α · a) · b + a · (α · b) , (1.4) a · (b · c) + b · (c · a) + c · (a · b) = 0 (1.5) where α, β, γ ∈ a0 and a, b, c ∈ a1. In particular, a0 is a (commutative) associative
subalgebra of a.
Weaker form of (1.3) The identity (1.3) needs a special discussion. This identity means that for every α ∈ a0 the operator 2 adα : a 7→ 2 α · a defines an action of the
commutative algebra a0 on the space a1. Furthermore, the identity (1.3) implies
Axioms of Lie antialgebra and derivations Recall that D ∈ End(a) is a derivation of a if for homogeneous x, y ∈ a (i.e., x, y are either purely even or purely odd) one has
D (x · y) = D (x) · y + (−1)p(D)p(x)x · D (y) . (1.7) This formula then extends by linearity for arbitrary x, y ∈ a. The space of all derivations of a is a Lie superalgebra denoted by Der(a).
Let us associate to every odd element a ∈ a1 the operator Ta : a → a of right
multi-plication by a
Ta(x) = x · a. (1.8)
The following observation partly clarifies the definition of Lie antialgebras.
The set of three identities: (1.6), (1.4) and (1.5) is equivalent to the condition that for all a ∈ a1 the operator Ta is a derivation.
Associativity of a0 The associativity axiom (1.2) seems quite different from the other
three axioms of Lie antialgebras. For instance, it has no interpretation in terms of deriva-tions. It turns out however, that this axiom can be understood as a corollary of the axioms (1.3)–(1.5).
Assume that every even element of a is a linear combination of products of odd ele-ments: α = Pai · aj. We will call such a Lie antialgebra ample. The following simple
statement is obtained in [7].
If a Lie antialgebra a is ample, then the identities (1.3), (1.4) and (1.5) imply (1.2).
Note that a similar property holds for Lie superalgebras.
Remark 1.1. It was noticed in [7] that Lie antialgebras are closely related to the Kaplan-sky superalgebras, see [8]. More precisely, a half-unital Lie antialgebra is a KaplanKaplan-sky superalgebra.
1.2
Examples
Let us give here a few examples of simple Lie antialgebras.
1. Our first example is a 3-dimensional Lie antialgebra called the tiny Kaplansky
superalgebra1 and denoted by K
3. This algebra has the basis {ε; a, b}, where ε is even and
a, b are odd, satisfying the relations ε · ε = ε
ε · a = 12a, ε · b = 12b, a · b = 12ε.
(1.9)
1
The algebra K3 is simple, i.e., it contains no non-trivial ideal. The corresponding algebra
of derivations is the simple Lie superalgebra osp(1|2), see [1] and Section 2.1. Moreover, this property completely characterizes the algebra K3.
Theorem 1. The algebra K3 is the unique finite-dimensional Z2-graded commutative
algebra such that the corresponding algebra of derivations is isomorphic to osp(1|2).
This theorem will be proved in Section 4.1.
2. The most interesting example of a Lie antialgebra we know is a simple infinite-dimensional Lie antialgebra with the basis εn, n ∈ Z; ai, i ∈ Z + 12
, where εn are
even and ai are odd and satisfy the following relations
εn· εm = εn+m,
εn· ai = 12an+i,
ai· aj = (j − i) εi+j.
(1.10)
This algebra is called the full derivation algebra2, see [8]. We will denote it by AK(1).
We will prove that AK(1) is closely related to the well-known Neveu-Schwarz conformal Lie superalgebra K(1) namely
K(1) = Der(AK(1)).
We conjecture that similarly to K3 the algebra AK(1) is the unique Z2-graded
commuta-tive algebra satisfying this property.
The Lie antialgebra AK(1) contains infinitely many copies of K3 with the basis
{ε0; ai, a−i} for each half-integer i.
3. The subalgebra of AK(1) with the basis n ε0, ε1, ε2, . . . ; a−1 2, a 1 2, a 3 2, . . . o
is also of some interest. This algebra is simple and can be understood as analog of the Lie algebra of (polynomial) vector fields on R. This algebra has interesting non-trivial cohomology studied in [6].
4. Let us finally explain how to construct a large class of examples of Lie antialgebras. Given an arbitrary commutative algebra C, there always exists an ample Lie antialgebra a such that a0 = C . An example is provided by K3(C) = C ⊗KK3(K). This already shows
that there are at least as many ample Lie antialgebras as there are commutative algebras. However, this is not the only possibility to realize a commutative algebra as an even par of a Lie antialgebra, so that there are much more Lie antialgebras than commutative algebras.
We will give more concrete examples of Lie antialgebras in Section 4.
2
1.3
The main properties
Lie antialgebras and Lie superalgebras and their representations Let V be a Z2-graded vector space. Consider the structure of Jordan Z2-graded algebra on End(V ) equipped with the anticommutator [A, B]+ = AB + (−1)p(a)p(b)BA. A representation of
a Lie superalgebra is a homomorphism
χ : a → (End(V ), [., .]+)
such that the image χ(a0) is commutative, i.e., χ(α) χ(β) = χ(β) χ(α) for all α, β ∈ a0.
This differs the notion of representation from the well-known notion of specialization of a Jordan superalgebra. Representations of K3 were studied in [9], study of representations
of AK(1) is an interesting problem, see Section 2.3 and 3.2 for some examples.
The most interesting property of Lie antialgebras is their relation to Lie superalgebras. With each Lie antialgebra a, we associate a Lie superalgebra ga in the following way.
The odd part (ga)1 coincides with a1 while the even part g0 is the symmetric tensor
square S2(a
1)a0, where the tensor product is defined over the commutative algebra a0.
Note that this construction is completely different from the classical Kantor-Koecher-Tits construction.
• Every representation of a Lie antialgebra a is a representation of the corresponding Lie superalgebra ga.
• The Lie superalgebra ga acts on a, in other words, there is a canonical
homomor-phism ga → Der(a).
One can say that the Lie antialgebra a selects a class of representations of ga that are
also representations of a. This is an interesting characteristic of representations of the Lie superalgebra ga.
The properties of the Lie superalgebra ga and the universal enveloping algebras U(ga)
and U(a) will be studied with more details in [7].
Odd “Lie-Poisson” bivector The notion of (odd) Lie-Poisson type bivector is the origin of Lie antialgebras, see Section 2.1. For an arbitrary Lie antialgebra a, the dual space with inverse parity, Π a∗, can be equipped with a canonical odd linear bivector
field Λa, see Section 3.3. Amazingly, the construction makes sense in the case of (purely
even) commutative associative algebra, i.e., for a = a0, but the dual space should be
understood as purely odd in this case. The bivector Λa is not Poisson in any sense. Its
general geometric characteristics is an interesting problem.
Central extensions and cohomology In Section 3.4, we introduce the notion of central extension of a Lie antialgebra. We prove that existence of the unit element ε ∈ a0
Classification In the finite-dimensional case, the classification of simple Lie antialge-bras is similar to the classification of commutative division algeantialge-bras, see Section 4.1. In the infinite-dimensional case, the situation is of course much more complicated. We also classify the Lie antialgebras of rank 1.
2
Lie antialgebras and symplectic geometry
In this section, we show the way Lie antialgebras appear in geometry. Notice that the invariant operations we construct are odd; we recover Lie antialgebra structures using the parity inversion functor.
2.1
The algebra K
3and the odd bivector Λ
Consider the vector space K2|1 equipped with the standard symplectic form, see [5],
ω = dp ∧ dq + 1
2dτ ∧ dτ, (2.1)
where p and q are the usual even coordinates on K2and τ is the formal Grassmann variable
so that τ2 = 0. An equivalent way to define this symplectic structure is to introduce the
Poisson bivector on K2|1: P = ∂ ∂p ∧ ∂ ∂q + 1 2 ∂ ∂τ ∧ ∂ ∂τ. (2.2)
which is inverse to the symplectic form: P = ω−1.
The Lie superalgebra osp(1|2) is defined as the space of infinitesimal linear transfor-mations preserving the symplectic structure. The bivector (2.2) is the unique (up to a multiplicative constant) even bivector invariant with respect to the action of osp(1|2). The odd bivector It turns out that there exists another, odd, osp(1|2)-invariant bivec-tor on K2|1.
Proposition 2.1. There exists a unique (up to a multiplicative constant) odd bivector
invariant with respect to the action of osp(1|2). It is give by the formula
Λ = ∂ ∂τ ∧ E + τ ∂ ∂p ∧ ∂ ∂q, (2.3) where E = p ∂ ∂p + q ∂ ∂q + τ ∂ ∂τ (2.4)
is the Euler field.
Proof. The osp(1|2)-invariance of Λ is a very easy check. Note that we will prove a much
Let us prove the uniqueness. An arbitrary odd bivector on K2|1 is given by Λ = ∂ ∂τ ∧ A + τ F ∂ ∂p ∧ ∂ ∂q,
where A is an even vector field and F is an even function. Let X be an even vector field, one has LXΛ = ∂ ∂τ ∧ [X, A] + τ X(F ) ∂ ∂p ∧ ∂ ∂q + τ F LX ∂ ∂p ∧ ∂ ∂q . If, furthermore, X ∈ osp(1|2), then it preserves the even part ∂
∂p ∧ ∂
∂q of the Poisson
bivector. The condition LXΛ = 0 then implies: [X, A] = 0 and X(F ) = 0.
The even part of osp(1|2) is a Lie algebra isomorphic to sl(2, K) and generated by the Hamiltonian vector fields with quadratic Hamiltonians hp2, pq, q2i. One checks that:
a) an even vector field A commuting with any even element of osp(1|2) is of the form A = c1τ
∂
∂τ + c2E, where c1 and c2 are arbitrary constants and E = p∂p∂ + q ∂q∂;
b) an even function F such that X(F ) = 0, where X is an even element of osp(1|2) is necessary a constant: F = c3.
The odd part of osp(1|2) is spanned by the following two vector fields: Xτ p = τ ∂ ∂q + p ∂ ∂τ, Xτ q = −τ ∂ ∂p + q ∂ ∂τ.
Applying any of these elements of osp(1|2) to the bivector Λ as above, one immediately gets c1 = c2 = c3.
The relation of the bivector Λ to the algebra K3 is as follows. Any bivector defines an
algebraic structure on the space of functions. Consider the bilinear operation associated with the bivector (2.3):
]F, G[ := (−1)
p(F )
2 hΛ, dF ∧ dGi, (2.5)
where F and G are arbitrary functions on K2|1, that is, F = F
0(p, q) + τ F1(p, q). This is
of course not a Poisson bracket.
Lemma 2.2. The space of linear functions on K2|1 equipped with the bracket (2.5) is a
Lie antialgebra isomorphic to K3.
Proof. One checks that after the parity inverting identification {ε; a, b} ←→ {τ ; p, q},
the algebra (1.9) coincide with the bracket (2.5) restricted to linear functions.
We proved that the Lie superalgebra osp(1|2) preserves the bivector Λ. Since osp(1|2) acts on K2|1 by linear vector fields, it also preserves the space of linear functions. In other
Remark 2.3. The bivector Λ given by (2.3), and the Lie antialgebra K3 are
equiva-lent structures, they contain the same information. The above lemma provides a parity inverting identification of the dual space:
Π K∗
3 ∼= (K2|1, Λ).
The bivector Λ is therefore analog of the “Lie-Poisson structure” corresponding to K3, cf.
Section 3.3 for a general setting.
An algebraic reformulation A purely algebraic way to reformulate the results of this section is as follows.
Consider the space of polynomials K[p, q, τ ] equipped with the standard action of the Lie supergroup OSp(1|2) (or, equivalently, of the Lie superalgebra osp(1|2)). We are looking for OSp(1|2)-invariant bilinear maps (., .) : K[p, q, τ ] ⊗ K[p, q, τ ] → K[p, q, τ ] satisfying the Leibniz rule in the both arguments, viz
(F G, H) = F (G, H) + (−1)p(G)p(H)(F, H) G and similarly in the second argument.
This problem has exactly two solutions.
1. The first operation is even, this is nothing but the standard Poisson bracket. It can be defined at order one by
{p, q} = 1, {p, τ } = 0, {q, τ } = 0, {τ, τ } = 1,
and then extended to K[p, q, τ ] via the Leibniz rule. Polynoms of order 1 span the Heisenberg Lie superalgebra.
2. The second operation is odd, it is defined at order 1 by
]p, q[ = 12τ, ]p, τ [ = 12p, ]q, τ [ = 12q, ]τ, τ [ = τ
and, again, extends to K[p, q, τ ] via the Leibniz rule. This is the “antibracket” (2.5), note that the homogeneous polynomials of order 1 form an algebra isomorphic to K3.
2.2
The full derivation algebra AK(1)
In this section we study the full derivation algebra AK(1) defined by formula (1.10). We show that AK(1) is simple and that Der(AK(1)) is isomorphic to the famous conformal Lie superalgebra K(1), also known as the (centerless) Neveu-Schwarz algebra.
We also prove that the conformal Lie superalgebra K(1) is the maximal algebra of vector fields on R2|1 that preserves the bivector (2.3). The algebra AK(1) can be viewed
as the maximal space of functions on R2|1 that form a Lie antialgebra with respect to the
The algebra AK(1) is simple We start with the following
Proposition 2.4. The relations (1.10) define a structure of a simple Lie antialgebra.
Proof. The identities (1.2) and (1.3) are evident. Let us prove the identity (1.4). One has
to check that
εn· (ai· aj) = (εn· ai) · aj + ai· (εn· aj)
One obtains in the left-hand-side 12 (j − i) εi+j+nand in the right-hand-side the sum of two
terms: 1
4 (j − (i + n)) εi+j+n and 1
4(j + n − i) εi+j+n, so that the identity (1.4) is satisfied.
Furthermore, the identity (1.5) reads:
(ai· aj) · ak+ (aj· ak) · ai+ (ak· ai) · aj = 0.
One obtains the sum 14(j − i) ai+j+n+14(k − j) ai+j+n+14(i − k) ai+j+n = 0. We proved
that AK(1) is, indeed, a Lie antialgebra.
It is quite easy to prove that AK(1) is simple, see, e.g. [8]. We do not dwell on the details here.
The conformal Lie superalgebra K(1) as the algebra of derivations The con-formal Lie superalgebra K(1) has the basis xn, n ∈ Z; ξi, i ∈ Z + 12
satisfying the following commutation relations
[xn, xm] = (m − n) xn+m, [xn, ξi] = i − n2 ξi+n, [ξi, ξj] = xi+j. (2.6)
It contains infinitely many copies of osp(1|2) with the generators {x−n, x0, xn; ξ−n 2, ξ
n 2}.
Define the following action of K(1) on AK(1): xn(ai) = i − n2 an+i, xn(εm) = m εn+m, ξi(aj) = (j − i) εi+j, ξi(εn) = ai+n. (2.7)
Note that his formula is well-known and represents the action of the algebra K(1) on the space of tensor densities of weight −12, see, e.g., [2, 3, 4] and Section 3.2.
Proposition 2.5. The action (2.7) preserves the structure of AK(1).
Proof. Consider for instance the action of an odd element of K(1). One has
together with
ξi(aj) · εk = (j − i) εi+j· εk= (j − i) εi+j+k
and
aj· ξi(εk) = aj · ai+k= 12(j − i − k) εi+j+k.
One finally gets:
ξi(aj · εk) = ξi(aj) · εk− aj · ξi(εk)
which is precisely the condition of odd derivation, see formula (1.7).
The action of other elements can be checked in the same way. Hence the result. We will prove in the end of this section that K(1) actually coincides with Der(AK(1)). Lie antialgebra AK(1) and symplectic geometry In this section we show that, similarly to K3, the Lie antialgebra AK(1) can be obtained from the odd Poisson
bivec-tor (2.3).
Consider the bracket (2.5) given by the explicit expression ]F, G[ = (−1) p(F ) 2 ∂F ∂τ E(G) − (−1) p(F )E(F )∂G ∂τ + τ ∂F ∂p ∂G ∂q − ∂F ∂q ∂G ∂p . (2.8) One checks that the full antialgebra of functions C∞(R2|1) equipped with this bracket is
not a Lie antialgebra.
Let Fλ be the space of homogeneous functions of degree −2λ on R2|1, that is, of the
functions satisfying the condition
E(F ) = −2λ F,
where E is the Euler field (2.4). We will allow F to have singularities in codimension 1, for instance, we can consider rational functions.
Note that there is a dense subspace of homogeneous functions on R2|1 that correspond
to the space of functions on R1|1. Indeed, given a function f (x, ξ) = f
0(x) + ξ f1(x) in one
even and one Grassmann variable, one defines a homogeneous of degree λ function (with singularities at (p = 0)) by Fλ f(p, q; τ ) = pλf q p, τ p . (2.9)
Proposition 2.6. The space, F−1
2, of homogeneous of degree 1 functions on R
2|1 is a Lie
antialgebra with respect to the antibracket (2.8) that contains AK(1).
Proof. The space of homogeneous of degree 1 functions is obviously stable with respect
to the antibracket (2.8) so that F−1
2 is an algebra. Define a bilinear operation f · g on the
One then easily checks the Lie antialgebra conditions. Choose the following “Taylor basis”:
ai = p q p i+1 2 , εn= τ q p n
and substitute it into the antibracket (2.5). One obtains the commutation relations (1.10), so that the Lie antialgebra AK(1) is a subalgebra of F−1
2.
The conformal Lie superalgebra K(1) also has a symplectic realization.
Proposition 2.7. The space F−1 of homogeneous of degree 2 functions on R2|1 is a Lie
superalgebra with respect to the Poisson bracket (2.2) that contains K(1).
Proof. The Poisson bracket of two homogeneous of degree 2 functions is, again, a
homo-geneous of degree 2 function. Therefore, F−1 is, indeed, a Lie superalgebra.
A homogeneous of degree 2 function can be written in the form (2.9) with λ = 2. Choosing the basis of the space of all such functions:
xn= p 2 2 q p n+1 , ξi = τ p q p i+1 2
and substituting it into the Poisson bracket (2.2), one immediately obtains the commu-tation relations (2.6). Therefore, K(1) is a subalgebra of F−1.
Remark 2.8. (a) The Lie superalgebra F−1 is a “geometric version” of the conformal Lie
superalgebra K(1), which is a polynomial part of F−1. Similarly, AK(1) is the polynomial
part of the Lie antialgebra F−1 2.
(b) The action (2.7) written in terms of homogeneous functions is, again, given by the standard Poisson bracket (2.2).
Lie superalgebra F−1 as algebra of symmetry
Proposition 2.9. The Lie superalgebra F−1 is the maximal Lie superalgebra of vector
fields that preserves the bivector (2.3).
Proof. Part 1. Let us first show that F−1 preserves the bivector (2.3). Given a function
H ∈ F−1, the corresponding Hamiltonian vector field is homogeneous of degree 0:
[E, XH] = 0. (2.11)
Consider a more general case, where P is a purely even (independent of τ ) Poisson bivector and E a purely even vector field. Assume P is homogeneous of degree −2 with respect to E, that is
LE(P ) = −2 P.
Let Λ be the odd bivector field
Λ = ∂
where
E = E + τ ∂ ∂τ. (Note that in our case E = p ∂
∂p + q ∂ ∂q, P = ∂ ∂p∧ ∂ ∂q.)
Let XH be a Hamiltonian (with respect to the Poisson structure P ) vector field
satis-fying the homogeneity condition (2.11). The Lie derivative of Λ along XH is as follows:
LXHΛ = LXH
∂ ∂τ
∧ E + XH(τ ) P + τ LXH (P ) .
If H is even, the above expression obviously vanishes. Consider now an odd function H = τ H1, then one gets from (2.2)
Xτ H1 = τ XH1 + H1
∂ ∂τ. Lemma 2.10. One has
LXτ H1(Λ) = hP ∧ E, dH1i (2.12)
where d is the de Rham differential. Proof. Using the obvious expressions
Xτ H1, ∂ ∂τ = XH1, LXτ H1(P ) = − ∂ ∂τ ∧ XH1, one obtains: LXτ H1(Λ) = XH1 ∧ E + H1P + τ ∂ ∂τ ∧ XH1 = XH1 ∧ E + H1P.
Finally, using the fact that E(H1) = H1, one obtains the expression (2.12).
The even tri-vector P ∧ E obviously vanishes on R2|1, and so we proved that X H,
indeed, preserves the bivector (2.3).
Part 2. Conversely, one has to show that any vector field preserving the bivector (2.3)
is a Hamiltonian vector field commuting with E. If X is a purely even vector field, i.e.,
X, ∂
∂τ
= 0 and X(τ ) = 0,
then LX(Λ) = 0 implies that X commutes with E and preserves the even bivector P = ∂
∂p ∧ ∂
∂q, so that X is Hamiltonian.
If X is an odd vector field:
X = F0
∂
∂τ + τ X0,
where F0 is an even function and X0 an even vector field, then one obtains explicitly
The assumption LX(Λ) = 0 implies that each of the three summands in this expression
vanishes. It follows from the equation E(F0) − F0 = 0, that F0 is a homogeneous of degree
1 function. The condition
XF0 + [E, X0] = 0
then implies that X0 is a vector field homogeneous of degree −1, since so is XF0, and thus
X0 = XF0. We proved that the vector field X is Hamiltonian and [E, X] = 0.
Proposition 2.9 is proved. Corollary 2.11. One has
F−1 = Der(F−1
2), K(1) = Der(AK(1)).
Indeed, the first statement is a reformulation of Proposition 2.7 while the second is its algebraic version. The subalgebra K(1) ⊂ F−1corresponds precisely to the space of vector
fields preserving the basis of AK(1) ⊂ F−1 2.
2.3
Representation of AK(1) by tangent vector fields
In this section we investigate the relation of the Lie antialgebra AK(1) to contact geom-etry. In some sense, AK(1) provides a way to “integrate” the contact structure.
The contact structure on S1|1 The natural projection R2|1\ {0} −→ S1|1, equips R1|1
with a structure of 1|1-dimensional contact manifold, see [2, 4, 3] for recent developments. This contact structure can be defined in terms of a contact 1-form α = dx + ξdξ, or, equivalently, in terms of an odd vector field3
D = 1 2 ∂ ∂ξ + ξ ∂ ∂x , since D spans the kernel of α.
The conformal Lie superalgebra K(1) can be realized as the Lie superalgebra of contact vector fields on S1|1. Every contact vector field on S1|1 is of the form
Xh = h(x, ξ)
∂
∂x + 2 D (h(x, ξ)) D,
where h(x, ξ) = h0(x) + ξ h1(x) is an arbitrary function. The map h 7→ Fh2, see (2.9),
is an isomorphism between Lie superalgebra of contact vector fields and F−1. The Lie
superalgebra of contact vector fields with polynomial coefficients is isomorphic to K(1).
3
Vector fields tangent to the contact distribution It turns out that there is a similar relation between the algebra AK(1) and the contact geometry.
A vector field tangent to the contact distribution is a vector field proportional to D, that is, X = f D for some function f (x, ξ).
Definition 2.12. We introduce the following anticommutator of tangent vector fields: [f D, g D]+:= f D ◦ g D + (−1)(p(f )+1)(p(g)+1)g D ◦ f D. (2.13) Note that the sign in this operation is inverse to that of usual commutator of vector fields. Remark 2.13. The space of tangent vector fields is not a Lie superalgebra since the Lie bracket of two tangent vector fields is not a tangent vector field, this is equivalent to non-integrability of the contact distribution.
Quite miraculously, that the anticommutator of two tangent vector fields is again a tangent vector field.
Proposition 2.14. The space of tangent vector fields equipped with the anticommutator
(2.13) is a Lie antialgebra that contains the Lie antialgebra AK(1). More precisely,
[χf, χg]+ = χf·g,
where f · g is the product (2.5).
Proof. Define a map from AK(1) to the space of tangent vector fields as follows. To each
homogeneous of degree 1 function F1
f(p, q, τ ), cf. formula (2.9), we associate a tangent
vector field by
χf = f (x, ξ) D. (2.14)
Let us calculate the explicit formula of the anticommutator (2.13).
Consider first the case where both of the functions, f and g, are odd, i.e., f = ξf1 and
g = ξg1. One then has
f D ◦ g D + g D ◦ f D = 1 2f g
∂
∂x + (f D(g) + g D(f )) D.
The first summand is zero since it contains ξ2 = 0, while the second one is equal to
ξ (f1g1) D, so that the anticommutator (2.13) on the odd functions coincides with f1g1.
This corresponds to the product f1 · g1 of two elements of the even part AK(1)0, see
formula (1.10).
If f = ξf1 is odd and g = g0 is even, then the Leibniz rule D ◦ f = D(f ) − f D implies
f D ◦ g D + g D ◦ f D = g D(f ) D = 12f1g0D,
so that one gets f1· g0 = 12f1g0, accordingly to the AK(1)0-action on AK(1)1, cf. (1.10).
If, finally, the both functions f and g are even, i.e., f = f0 and g = g0, then
f D ◦ g D − g D ◦ f D = (f D(g) − g D(f )) D = 1 2ξ (f0g
′
0− g0f0′) D,
gives the skew-symmetric product f0 · g0 = 12 (f0g′0− g0f0′) , on AK(1)1 with values in
Remark 2.15. The map (2.14) is nothing but the bivector (2.3) contracted with the elements of AK(1). One checks that
χf(g) = (−1)p(f ) 2 Λ, dFf1∧ dFg0 , (2.15) where F1
f and Fg0 are functions on R2|1 homogeneous of degree 1 and 0, respectively,
obtained as the lift of f and g according to (2.9). It is interesting to compare the above formula with (2.10).
We conclude this section with the formula for the product on the space of smooth functions on S1|1 that coincides with (1.10) on the polynomial basis. The functions are
of the form f (x, ξ) = α(x) + ξ a(x), the product is given by (α + ξ a) · (β + ξ b) = αβ + ab′ − a′b + 1
2ξ (αb + βa) . (2.16)
This is the C∞-analog of AK(1).
2.4
A pair of symplectic forms on R
4|2and the algebras K
3(C)
and AK(1)
CConsider the space K2n|m with linear symplectic form. There is no analog of the odd
bivector (2.3), if n ≥ 2 or m ≥ 2, since the only OSp(m|2n)-invariant bivector is the Poisson bivector. In this section we investigate the second special case: n = m = 2. It turns out that one has to fix a pair of symplectic forms and consider the group of linear transformations preserving both of them.
Note that, if n ≥ 3 or m ≥ 3, then one has to consider degenerate linear 2-forms, cf. Section 3.3.
A pair of symplectic forms on K4|2 Consider two generic linear symplectic forms
on K4|2. There exist linear coordinates (p
1, q1, p2, q2; τ1, τ2) such that these forms are as
follows.
ωε= dp1∧ dq1+ dp2∧ dq2+ 12(dτ1 ∧ dτ1− dτ2∧ dτ2)
and
ωσ = dp1∧ dp2− dq1∧ dq2+ dτ1∧ dτ2.
This is a very simple fact of linear algebra.
Note that, in the real case, the forms ωε and ωσ are, respectively, the real and the
imaginary parts of the complex 2-form (2.1), where p, q; τ are the following complex coordinates p = p1+ iq2, q = q1+ ip2, τ = τ1+ iτ2.
Proposition 2.16. The Lie superalgebra of linear vector fields on K4|2 preserving the two
symplectic forms ωε and ωσ is isomorphic to:
(i) osp(1|2, C), if K = R;
Proof. The Lie superalgebra of linear vector fields on K4|2 preserving the two symplectic forms ωε and ωσ is a subalgebra of osp(2|4, K) spanned by 6 even Hamiltonian vector
fields corresponding to the quadratic Hamiltonians
p2
1− q22, p22− q12, p1p2+ q1q2, p1q1 − p2q2, p1q2, q1p2
and 4 odd bi-Hamiltonian vector fields with the Hamiltonians
{p1τ1− q2τ2, p2τ1+ q1τ2, q1τ1− p2τ2, q2τ1 + p1τ2} .
In the real case, this defines an osp(1|2, C)-action on R4|2. In the complex case, this is
osp(1|2, C)C∼= osp(1|2, C) ⊕ osp(1|2, C).
Remark 2.17. Note that, in the case (i), osp(1|2, C) is viewed as a simple Lie superal-gebra over R. This alsuperal-gebra of derivations is thus obtained as intersection of two copies of the real symplectic Lie superalgebra:
osp(1|2, C) = ospε(1, 1 | 4) ∩ ospσ(1, 1 | 4)
corresponding to the symplectic forms ωε and ωσ, respectively.
The bivector ΛC The following bivector is the unique (up to a multiplicative constant)
odd osp(1|2, C)-invariant bivector on K4|2:
ΛC = ∂ ∂τ1 ∧ E + ∂ ∂τ2 ∧ J + τ1πε+ τ2πσ, (2.17) where E = X i=1,2 pi ∂ ∂pi + qi ∂ ∂qi + τi ∂ ∂τi
is the Euler vector field and J = q2 ∂ ∂p1 + p2 ∂ ∂q1 − q1 ∂ ∂p2 − p1 ∂ ∂q2 + τ2 ∂ ∂τ1 − τ1 ∂ ∂τ2 (2.18) and where the bivectors πε and πσ are
πε = ∂ ∂p1 ∧ ∂ ∂q1 + ∂ ∂p2 ∧ ∂ ∂q2 , πσ = ∂ ∂p1 ∧ ∂ ∂p2 − ∂ ∂q1 ∧ ∂ ∂q2 .
It is now easy to check that the space of linear functions on (R4|2, ΛC) form a Lie
Algebra AK(1)C and the pair of symplectic structures on R4|2 Let us now realize
the algebras AK(1)Cand K(1)Cin terms of real rational harmonic functions on R4|2. This
is related to the bi-Hamiltonian formalism defined by the pair of symplectic structures ωε
and ωσ.
The Lie antialgebra AK(1)C is represented by the homogeneous of degree 1 harmonic
functions on R4|2 with the complex structure (2.18). The odd generators of AK(1) C are as follows εn= 21nRe τ q p n , βn= 21nIm τ q p n , and the even ones are
an = 21nRe p q p n , bn = 21n Im p q p n ,
where p, q, τ are the complex coordinates. One then checks the relations in AK(1)C.
The conformal Lie superalgebra K(1)Cis realized by homogeneous bi-Hamiltonian
vec-tor fields on R4|2. For each function F ∈ C∞(R4|2), denote by Xε
F and XFσ the Hamiltonian
vector fields on R4|2 with respect to the symplectic form ω
ε and ωσ, respectively.
One checks that the following three conditions are equivalent. 1. F is a homogeneous of degree 2 harmonic function:
F = Re p2fq p, τ p , where f is an arbitrary function.
2. The function F satisfies the relations: E(F ) = 2F and J (J (F )) = −4F . 3. The Hamiltonian vector fields with the Hamiltonian F commute with E and J :
[E, Xε F] = [E, X σ F] = [J , X ε F] = [J , X σ F] = 0
and are bi-Hamiltonian such that Xε F = −12X σ J (F ), XFσ = 12X ε J (F ).
This space of homogeneous harmonic bi-Hamiltonian vector fields is a Lie superalgebra isomorphic to K(1)C.
Proposition 2.18. The Lie superalgebra K(1)C is the maximal algebra of vector fields on
R4|2 preserving the bivector (2.17).
Proof. Proposition 2.9 implies that K(1)C, indeed, preserves the bivector ΛC. The proof
3
Elements of the general theory
In this section we investigate the relations between Lie antialgebras and Lie superalgebras. We introduce the notion of representation and module over a Lie antialgebra and show how to extend them to the corresponding Lie superalgebra.
We also define the analog of the Lie-Poisson structure for an arbitrary Lie antialgebra a. This is an odd linear bivector field on the dual space with inverse parity Π a∗. This notion
relates Lie antialgebras to geometry.
We finally define and study the notion of central extensions that will later be useful for classification results.
3.1
Relation to Lie superalgebras
In this section we associate a Lie superalgebra ga with an arbitrary Lie antialgebra a. The
construction will be important since representations of a Lie antialgebra can always be extended to the corresponding Lie superalgebra.
Definition 3.1. Consider the space ga = (ga)0⊕ (ga)1 defined as follows. The even part,
(ga)0, is the symmetric tensor square of a1 over a0:
(ga)0 := a1⊙a0 a1,
more precisely,
(ga)0 = {(a ⊗ b + b ⊗ a)/∼ | a, b ∈ a1} ,
where the equivalence relation ∼ is defined by
α · a ⊗ b ∼ a ⊗ α · b
for all a, b ∈ a1 and α ∈ a0. The odd part of (ga), is nothing but the odd part of a:
g1 := a1.
The Lie bracket on ga is defined by
[a ⊙ b , c ⊙ d] = a · (b · c) ⊙ d + b · (a · c) ⊙ d − c · (d · a) ⊙ b − d · (c · a) ⊙ b, [a ⊙ b , c] = a · (b · c) + b · (a · c) ,
[a , b] = a ⊙ b.
(3.1)
Theorem 2. The space ga endowed with the bracket (3.1) is a Lie superalgebra.
The action of gaon a Let us define a Lie superalgebra homomorphism T : ga→ Der(a).
The restriction T (ga)1 is naturally defined since the odd part (ga)1 is identified with a1
and the odd part a1 acts on a by derivations, cf. formula (1.8). One then extends this
map to the even part (ga)0 in a unique way as follows: Ta⊙b := [Ta, Tb]. One then has
explicitly
Ta⊙bx = (x · a) · b + (x · b) · a (3.2)
for odd a, b ∈ a1 and arbitrary x ∈ a.
Lemma 3.2. (i) The map (3.2) is well-defined, that is, it is compatible with the
equiva-lence relation ∼.
(ii) For all a, b ∈ a1, the linear map Ta⊙b ∈ End(a) is an even derivation.
Proof. Part (i). The computation is straightforward but we provide here some details in
order to show how work the axioms of Lie antialgebra. One has to check that
(x · (α · a)) · b + (x · b) · (α · a) = (x · (α · b)) · a + (x · a) · (α · b)
for every α ∈ a0 and arbitrary x. Consider, for instance, the case where x is odd, that is
x ∈ a1. Using (1.3), one obtains
(x · b) · (α · a) = α · ((x · b) · a) while using (1.4) and then (1.5) and again (1.3), one obtains
(x · (α · a)) · b = (α · (x · a)) · b − ((α · x) · a) · b
= 2 α · ((x · a) · b) + (a · b) · (α · x) + (b · (α · x)) · a for the terms in the left-hand-side. Similarly, for the right-hand-side one has
(x · a) · (α · b) = α · ((x · a) · b)
(x · (α · b)) · a = 2 α · ((x · b) · a) − ((α · x) · b) · a. Finally, collecting the terms in the (lhs) − (rhs) one gets
α · ((x · a) · b) − α · ((x · b) · a) + (a · b) · (α · x) = −α · ((a · b) · x) + (a · b) · (α · x) = 0, using (1.5).
The computation for an even element x ∈ a0 is similar.
Part (ii) is obvious since Ta⊗b is a commutator of two derivations.
Example 3.4. In the case of the simple algebras a = K3 and AK(1), the corresponding
Lie superalgebras ga coincide with the respective algebra of derivations:
gK3 = osp(1|2), gAK(1) = K(1).
However, in general this is not the case. The more Lie antialgebra a is far of being simple, the more the corresponding Lie superalgebra gl(a) is far of Der(a).
3.2
Representations and modules
In this section we introduce important notions of representation and of module over Lie antialgebras. Amazingly, these two notions are different for Lie antialgebras. We will investigate these notions in some details for the algebras K3(C) and AK(1).
Representations: the definition Let V = V0 ⊕ V1 be a Z2-graded vector space.
Consider the Jordan algebra structure on the space End(V ) defined by the anticommutator [A, B]+ = A B + (−1)p(A)p(B)B A. (3.3) Note that this operation the sign rule inverse to that of the usual Z2-graded commutator
(the sign is “−” if and only if both A and B are odd).
Remark 3.5. Note that the Jordan algebra (End(V ), [., .]+) is of course not a Lie
antial-gebra. However, we use this algebra to define the notion of representation.
Given a Lie antialgebra a, an (even) linear map χ : a → End(V ) is called a
represen-tation of a, if two conditions are satisfied:
χx·y = [χx, χy]+, (3.4)
for all x, y ∈ a and the images of even elements are commuting operators:
χαχβ = χβχα (3.5)
for all α, β ∈ a0.
Remark 3.6. A linear map χ : a → End(V ) satisfying (3.4), but not necessarily (3.5) is called a specialization of a Jordan algebra. This means, a representation is a very particular case of specialization. The condition (3.5) is crucial for us.
Example 3.7. We have already considered a particular case of the anticommutator (3.3), namely the operation (2.13) on the space of vector fields on S1|1 tangent to the contact
structure. The map (2.14) defines a representation of the full derivation algebra AK(1) in the space of differential operators on R1|1.
Theorem 3. Every representation of a Lie antialgebra a is naturally a representation of
Proof. Given a representation χ of a Lie antialgebra a, the construction of the
correspond-ing representation of ga is as follows. The odd part (ga)1 coincides with a1, so that the
map χ|(ga)1 is already defined. For the even elements of ga we take the usual commutator:
χa⊙b := [χa, χb] = χa◦ χb+ χb◦ χa.
To prove Theorem 3, one has to show that this is indeed a representation of ga.
The complete proof is again a very complicated but straightforward computation, that we omit. The details are given in [7].
The above result shows that representations of a is some particular class of represen-tations of ga.
Representations of K3 Representations of the simple algebra K3 were studied in [9].
The corresponding Lie superalgebra is osp(1|2), so that representations of K3 is a
particular class of representations of osp(1|2). It turns out that this class is characterized in terms of the classical Casimir element C ∈ U(osp(1|2)). The following statement is the main result of [9].
There is a one-to-one correspondence between representations of s Lie antialgebra K3
and the representations of the Lie superalgebra osp(1|2) such that χ(C) = 0.
A representation of K3 is obviously given by one even operator E ∈ End(V ) and two
odd operators A, B ∈ End(V ) satisfying the relations AB − BA = E, AE + EA = A, BE + EB = B, E2 = E.
(3.6)
Among other classification results of [9], let us mention the following simple but beautiful statement.
Up to isomorphism, the operator E in (3.6) is of the form:
E|V0 = 0, E|V1 = Id. (3.7)
Example: tensor-density representations of K(1) and AK(1) Consider the repre-sentations of K(1) called tensor density reprerepre-sentations Fλ, where λ ∈ C is the parameter.
The basis in Fλ is {fn, n ∈ Z, φi, i ∈ Z + 12} and the action of K(1) is given by
χxn(fm) = (m + λn) fn+m, χxn(φi) = i + (λ + 1 2)n φn+i, χξi(fn) = n 2 + λi φi+n, χξi(φj) = fi+j.
For instance, the adjoint representation of K(1) is isomorphic to F−1. The K(1)-action
(2.7) on AK(1) is isomorphic to F−1
2 with inverse parity.
It is now very easy to check that the K(1)-module Fλ is a representation of AK(1) if
and only if
λ = 0 or λ = 1 2. Note that the modules F0 and F1
2 are dual to each other, these two modules are known
to be special.
Modules over Lie antialgebras: the definition A Z2-graded vector space V is called
an a-module if there is an even linear map ρ : a → End(V ), such that the direct sum a⊕ V equipped with the product
(a, v) · (b, w) = a · b , ρaw + (−1)p(b)p(v)ρbv
, (3.8)
where a, b ∈ a and v, w ∈ V are homogeneous elements, is again a Lie antialgebra. We call the Lie antialgebra structure (3.8) a semi-direct product and denote it by a ⋉ V . Example 3.8. 1). The “adjoint action” ad : a → End(a) defined by
adxy = x · y,
for all x, y ∈ a, defines a structure of a-module on a itself. This follows, for instance, from the fact that the tensor product C ⊗ a of a Lie antialgebra a with a commutative algebra C is again a Lie antialgebra. Indeed, consider C = K[t]/(t2), then one has C ⊗ a = a ⋉ a.
2). The “coadjoint action” ad : a → End(a∗) defined by
adxϕ = (−1)p(x)p(ϕ)ad∗xϕ,
for x ∈ a and ϕ ∈ a∗, defines a structure of a-module on a∗.
3.3
The odd Lie-Poisson type bivector
In this section we introduce the notion of canonical odd bivector field Λa on the dual
space Π a∗.
The rank and the pencil of presymplectic forms Given a Lie antialgebra a, we call the dimension of the odd part a0 of a Lie antialgebra a the rank of a:
rk a := dim a0.
Let a be a Lie antialgebra of rank r. Fix an arbitrary basis {ε1, . . . , εr} of a0. One
obtains a set of r bilinear skew-symmetric (or presymplectic) forms: {ω1, . . . , ωr} on a1
by projection on the basic elements: a · b =: r X i=1 ωi(a, b) εi, (3.9) for all a, b ∈ a1.
Changing the basis, one obtains linear changes of the corresponding set of skew-symmetric forms. Therefore, the pencil of presymplectic forms hω1, . . . , ωri is well-defined.
To each 2-form ωi, one associates a bivector πi ∈ ∧2a∗1, that we can understand as a
con-stant bivector field on a∗ 1.
The adjoint vector fields To each element α ∈ a0, one associates an even linear
operator Aα : a → a defined by
Aα|a0 = adα, Aα|a1 = 2 adα. (3.10)
These linear operators can be, of course, viewed as linear vector fields on a∗.
We will denote A1, . . . , Ar the (even) vector fields on a∗ corresponding to the elements
of the basis {ε1, . . . , εr}.
Example 3.9. In the important case where a0 contains the unit element ε and the center
of a is trivial (see Theorem 4 below), one has Aε = E, where E is the Euler vector field
on the vector space a∗, i.e., the generator of the K∗-action by homotheties.
The definition We will use the parity inversion functor Π. Consider the space Πa∗ 0
and denote by (τ1, . . . , τr) the Grassmann coordinates dual to the chosen basis. Choose
furthermore arbitrary linear coordinates (x1, . . . , xn) on Πa∗1.
Define the following odd bivector on Πa∗:
Λa= r X i=1 ∂ ∂τi ∧ Ai+ τiπi . (3.11)
The corresponding antibracket on the space of (polynomial, smooth, etc.) functions on Πa∗ is defined as in (2.5). This antibracket is obviously linear, i.e., the space of linear
Proposition 3.10. Linear functions on the space (Πa∗, Λa) form a Lie antialgebra
iso-morphic to a.
Proof. The antibracket of two even linear functions obviously corresponds to (3.9). The
odd linear functions on Πa∗ are linear combinations of τ
1, . . . , τr. The antibracket of
an odd and an even linear functions is given by ]τi, ℓ[= 12Ai(ℓ), where ℓ ∈ a0. This
corresponds precisely to the adjoint action of εi on ℓ.
Finally, the antibracket of two odd linear functions is given by ]τi, τj[= 12(Ai(τj) + Aj(τi)) ↔ 12 adεiεj+ adεjεi
= εi· εj.
Hence the result.
Corollary 3.11. The bivector (3.11) is independent of the choice of the basis.
Example 3.12. The bivectors (2.3) and (2.17) are precisely the canonical bivectors on Π K3(R)∗ and Π K3(C)∗, respectively.
Example 3.13. The bivector (2.3) makes sense in the case of a (purely even) commutative algebra, i.e., where a1 = {0}. Let C be a commutative associative algebra with basis
{ε1, . . . , εr} and structural constants ckij such that εi· εj = Pkckijεk. Let τ1, . . . , τr be a
set of Grassmann coordinates. The bivector (3.11) is then of the form ΛC = X i,j,k ciijτk ∂ ∂τi ∧ ∂ ∂τj .
We understand this linear bivector field as commutative analog of the Lie-Poisson bivector. The conformal Lie superalgebra The Lie superalgebra Der(a) can be viewed as the algebra of linear vector fields on Πa∗ preserving the bivector Λ
a. It is natural to define
(an infinite-dimensional) Lie superalgebra of “conformal derivations” of a as the algebra of vector fields on Πa∗ preserving the bivector Λ
a. We simply drop the linearity condition.
Example 3.14. The algebra K(1) is the algebra of conformal derivations of K3.
The notion of algebra of conformal derivations of a deserves a special study. It would be also interesting to understand if there is a general notion of conformal Lie antialgebra. It makes sense to look for such a notion in terms of the algebra of functions on Πa∗
homogeneous with respect to the vector fields A1, . . . , Ar.
3.4
Central extensions
In this section we define the notion of extension of a Lie antialgebra a with coefficients in any a-module. It will be useful for the classification result of Section 4.2. The notion of extension is a part of a general cohomology theory that will be developed in [6].
Definition 3.15. (a) An exact sequence of Lie antialgebras
0 −−−→ V −−−→ ea −−−→ a −−−→ 0 (3.12)
is called an abelian extension of the Lie antialgebra a with coefficients in V . As a vector space, ea = a ⊕ V , and the subspace V is obviously an a-module.
(b) An extension (3.12) is called non-trivial if the Lie antialgebraea is not isomorphic to the semi-direct sum a ⋉ V .
(c) If the subspace V belongs to the center ofea, then the extension (3.12) is called a
central extension.
In this section we develop the notion of central extension. Since any central extension can be obtained by iteration of one-dimensional central extensions, it suffice to consider only the case of one-dimensional central extensions. One then has two possibilities:
dim V = 0|1, or dim V = 1|0.
We then say that the one-dimensional central extension is of type I or of type II, respec-tively.
Central extensions of type I The general form of central extensions of type I is as follows.
Proposition 3.16. (i) A central extension of type I is defined by (an even) symmetric
bilinear map C : a ⊗ a → K0|1 satisfying the the following conditions:
C (α, β · a) = C (β, α · a) = 12 C (α · β, a)
C (a · b, c) + C (b · c, a) + C (c · a, b) = 0, (3.13)
for all α, β ∈ a0 and a, b, c ∈ a1.
(ii) A central extension is trivial if and only if there exists an even linear function
f : a0 → K such that
C(α, a) = f (α · a) (3.14)
for all α ∈ a1 and a ∈ a0.
Proof. Part (i). Given a map C as in (3.13), let us define an antibracket on a ⊕ K. We
fix an element z ∈ K and set:
x · y = (x · y)a+ C(x, y) z, x · z = 0, (3.15)
for all x, y ∈ a. Note that, since the map C is symmetric and even, one has
C(x, y) = C (x1, y0) + C (y1, x0) , (3.16)
where x1, y1 ∈ a1 and x0, y0 ∈ a0. One then easily checks that formula (3.15) defines a
Conversely, a Lie antialgebra structure on a ⊕ K0|1 such that the subspace K0|1 belongs to the center is obviously of the form (3.15).
Part (ii). In the case where C is as in (3.14), the linear map a ⊕ K0|1 → a ⊕ K0|1
given by (x, z) 7→ (x, z + f (x)) intertwines the structure (3.15) with the trivial direct sum structure. This means that the central extension is trivial.
Conversely, if the extension is trivial, then there exists an intertwining map a ⊕K0|1 →
a⊕ K0|1 sending the structure (3.15) to the trivial one. This map can, again, be chosen
in the form (x, z) 7→ (x, z + f (x)), since a different choice of the embedding of K0|1 does
not change the structure.
We will call a map C satisfying (3.13) a 2-cocycle of type I. A 2-cocycle of the form (3.14) will be called a coboundary.
Central extensions of type II
Proposition 3.17. (i) A central extension of type II is defined by an even symmetric
bilinear map C : a ⊗ a → K satisfying the following identities:
C (α, a · b) = C (α · a, b) + C (a, α · b)
C (α · β, γ) = C (α, β · γ) , (3.17)
for all α, β, γ ∈ a0 and a, b ∈ a1.
(ii) The extension is trivial if and only if there exists an even linear functional f :
a1 → K such that
C(x, y) = f (x · y), (3.18)
for all x, y ∈ a.
Proof. The proof is similar to that of Proposition 3.16.
We will call an odd map (3.17) satisfying (3.17) a 2-cocycle of type II. In the case where it is given by (3.18), the map C is called a coboundary.
Example 3.18. Consider the kernel of the presymplectic pencil of a: I =
r
\
i=1
ker ωi,
where r = rk a, the forms ωi are defined by (3.9). In other words,
I = {a ∈ a1 | a · b = 0, for all b ∈ a1}.
Proof. By definition, I is an abelian subalgebra and the bracket of a ∈ I with any element
b ∈ a1 vanishes. One has to show that α · a ∈ I, for arbitrary α ∈ a0 and a ∈ I. Indeed,
using the identity (1.4), one obtains
(α · a) · b = α · (a · b) − a · (α · b) = 0 since for a ∈ I and every b ∈ a1, one has a · b = 0.
It follows that the Lie antialgebra a itself is an abelian extension of type I of the quotient-algebra a/I:
0 −−−→ I −−−→ a −−−→ a/I −−−→ 0. Let us also outline the case where this extension has to be central.
Proposition 3.20. If a is ample, then the ideal I belongs to the center of a.
Proof. The ideal I belongs to the center of a if and only if the action of a0 on I is trivial.
Recall that a is ample if map a1⊗ a1 → a0 is surjective. Surjectivity means that for every
α ∈ a0 there are a, b ∈ a1 such that α = a · b. Using the Jacobi identity (1.5), one obtains
for every c ∈ I:
α · c = (a · b) · c = − (b · c) · a − (c · a) · b = 0, since both summands in the right-hand-side vanish.
The Lie antialgebra a is therefore a central extension of a/I (of type I).
The case of the unit element Let us consider the case where the associative com-mutative algebra a0 contains the unit element ε, and show the Lie antialgebra a has
essentially no non-trivial central extensions in this case.
Theorem 4. If the Lie antialgebra a contains the unit element ε ∈ a0, then a is a direct
sum:
a= a ⊕ Z(a), (3.19)
of its center and the Lie antialgebra a = a/Z(a) that has no non-trivial central extensions. Proof. Let us consider the action of ε on a1. The identity (1.3) implies the “half-projector”
relation: ad1ε ◦ ad1ε = 12ad 1 ε. Therefore, a1 is split a1 = a1,1 2 ⊕ a1,0 to a sum of 1
2- and 0-eigenspaces, respectively. That
is, ade|a1,12
= 12Id and ade|a1,0 = 0.
Proof. Let a ∈ a1,0, that is, ε · a = 0. One has to show that x · a = 0 for all x ∈ a.
Let first x = α be an element of a0. The identity (1.3) implies
ε · (α · a) = ε · (α · a) + α · (ε · a) = (e · α) · a = α · a. But then ad1ε◦ ad1ε = 12ad
1
ε implies α · a = 0.
In the case where x = b is an element of a0 the proof is similar.
Let us show now that a has no non-trivial central extensions.
Let C be a 2-cocycle of type I on a. Apply the first identity (3.13) to β = ε, where ε is the unit. One has
C(α, a) − C(α, ε · a) = C(ε, α · a).
If a ∈ Z(a), this formula implies C(α, a) = 0. If a is an element of the 1
2-eigenspace a1,1 2
of the unit element ε, then one obtains C(α, a) = 2 C(ε, α · a). Therefore, the cocycle C is a coboundary.
Let now C be a 2-cocycle of type II. It can be decomposed into a pair (C0, C1) of maps
C0 : a0⊗ a0 → K, C1 : a1⊗ a1 → K,
where C0 is symmetric and C1 is skew-symmetric. The first condition (3.17) gives
C0(ε, a · b[) = C1(ε · a, b) + C1(a, ε · b),
so that C1(a, b) = C0(ε, ]a, b[). The second condition (3.17) implies C0(α, β) = C0(ε, α·β).
Therefore, the cocycle C is, again, a coboundary. Proof of Theorem 4 is complete.
4
Classification results
In this section, we prove that K3(C) is the only finite-dimensional complex simple Lie
antialgebra. In the real case one has K3(R) and K3(C). This means that the situation is
similar to the case of commutative algebras. We also prove that K3 is characterized by
the fact that its algebra of derivations is isomorphic to osp(1|2).
We also obtain a complete classification of finite-dimensional Lie antialgebras of rank 1, i.e., with dim a0 = 1. This, in particular, provides with a number of examples of Lie
antialgebras, other that we already considered.
4.1
Classification of finite-dimensional simple Lie antialgebras
We call a Lie antialgebra a simple if it contains no ideal except for the trivial one and a itself. The classification of finite-dimensional simple Lie antialgebras is similar to that of commutative algebras.
Theorem 5. (i) There exists a unique finite-dimensional complex simple Lie antialgebra.
Recall that the only simple finite-dimensional commutative algebras over R are R and C= R + iR. In the complex case there is only C itself.
Let us start with the complex case. Let a be a simple finite-dimensional Lie antialgebra. We will first assume that the commutative algebra a0 has no nilpotent elements. As it is
very well known, a0 is of the form
a0 = C ⊕ · · · ⊕ C| {z } rtimes
.
We will prove that if a is simple, then r = 1.
Choose a basis {ε1, . . . , εr} in a0 such that εi·εj = δij. As in Section 3.3, one associates
with each element εi a presymplectic form ωi on a1 (see formula 3.9).
The following statement shows that, if r > 1, then the algebra a contains a non-trivial ideal. Consider the subspace ker ω1 ⊂ a1 consisting of the elements a ∈ a1 such that, for
all b one has: a · b is a combination of εi with i ≥ 2. Consider also the following subspace
of a:
I = ker ω1⊕ hε2, · · · , εri.
Lemma 4.1. The subspace I is an ideal of a.
Proof. One has to prove that
(a) α · I ⊂ I, for an arbitrary α ∈ a0;
(b) a · I ⊂ I, for an arbitrary a ∈ a1.
Part (a). Let a ∈ ker ω1, by identity (1.4), one has for an arbitrary b ∈ a1:
(α · a) · b = α · (a · b) − a · (α · b) .
The both terms in the right-hand-side are combinations of εi with i ≥ 2. Therefore, one
obtains α · a ∈ ker ω1.
Part (b). Let α ∈ hε2, · · · , εri, then ε1·α = 0. Let a ∈ a1be arbitrary, one has to prove
that, again, α · a ∈ ker ω1. Choose an arbitrary b ∈ a1 and consider again the expression
(α · a) · b. Since this is an even element of a, it can be written in the form: (α · a) · b =
r
X
i=1
ciεi.
One has to prove that c1 = 0. One has ε1· ((α · a) · b) = (ε1 · (α · a)) · b + (α · a) · (ε1· b) .
But the first summand in the right-hand-side vanishes. Indeed, by (1.3), one has ε1· (α · a) = 12(ε1· α) · a = 0,
since ε1· α = 0. In the same way, one obtains
However, for the left-hand-side, one obtains using (1.2):
ε1· (ε1· ((α · a) · b)) = (ε1· ε1) · ((α · a) · b) = ε1· ((α · a) · b) = c1
since ε1· ε1 = ε1; while, for the right-hand-side, one gets using (1.3):
(α · a) · (ε1· (ε1· b)) = 12(α · a) · ((ε1· ε1) · b) = 12(α · a) · (ε1· b) = ε1· ((α · a) · b) = 12c1.
Therefore, c1 = 0 and so α · a ∈ ker ω1. The result follows.
Lemma 4.1 implies that dim(a0) = 1. Choose a non-zero element ε ∈ a0 and denote ω
the corresponding 2-form.
Lemma 4.2. The form ω is of rank 2.
Proof. Choose a canonical (Darboux) basis {a1. . . , an, b1, . . . , bn, c1, . . . , cm}, so that
ω(ak, bℓ) = δkℓ, ω(ak, aℓ) = ω(bk, bℓ) = ω(ck, .) = 0.
where k, ℓ = 1, . . . , n. Let us show that n = 1. First, n ≥ 1, since otherwise ω = 0 and a1
is an ideal. It follows that there are some elements ak and bk such that ε = ak· bk.
Assume that n > 1. The identity (1.5) implies
ε · aℓ = (ak· bk) · aℓ = − (bk· aℓ) · ak− (aℓ· ak) · bk = 0
for any k 6= ℓ. It follows that ε · a = 0, for any a ∈ a1. Furthermore,
ε · ε = ε · (ak· bk) = (ε · ak) · bk+ ak· (ε · bk) = 0,
for any i = 1, . . . , n. Therefore, ε belongs to the center of a; in particular, a cannot be simple. This is a contradiction.
Lemma 4.1 and Lemma 4.2 imply Theorem 5 in the complex case, where the commu-tative algebra a0 has no nilpotent elements. If now a0 = Cn⋉N , where N is a nilpotent
ideal, then the same arguments prove that n ≤ 1 and ker ω1⊕ N is an ideal. Theorem 5
is proved in the complex case.
The real case immediately follows from the complex one. Indeed, let a be a real simple Lie antialgebra, the standard arguments show that the complexification a ⊗RC is either
simple or the direct sum of two isomorphic simple ideals. Theorem 5 is proved.
Proof of Theorem 1 Let a be a finite-dimensional commutative Z2-graded algebra such
that Der(a) ∼= osp(1|2). As an osp(1|2)-module, a is a direct sum of irreducible modules. Recall that finite-dimensional irreducible osp(1|2)-modules are the modules D(h/2) with positive integer h. These modules are of dimension 2h + 1 and of highest weight h. It follows that for two elements of a such that x1 ∈ D(h1/2) and x2 ∈ D(h2/2), one has:
Since a is finite-dimensional, there exists a non-zero submodule D(h′/2) ⊂ a with maxi-mal h′.
If h′ 6= 0, then (4.1) implies that D(h′/2) belongs to the center of Z(a). However, the
Lie superalgebra End(Z(a)) is a subalgebra of Der(a) so that Der(a) cannot be isomorphic to osp(1|2) if a has a non-trivial center.
Theorem 1 is proved.
4.2
Lie antialgebras of rank 1
Let us assume that the commutative algebra a0is one-dimensional. There are two different
possibilities:
1. a1 = K, as a commutative algebra, so that it contains the unit element denoted by
ε, such that ε · ε = ε;
2. a1 is nilpotent and the only odd generator α satisfies α · α = 0.
The structure of the algebra a is characterized by one bilinear skew-symmetric form on a1 defined by
a · b = ω(a, b) α,
where α is a (unique up to a constant) non-zero element of a1.
Let us first construct several examples of Lie antialgebras of rank 1.
A. The form ω is non-degenerate (A1) The (2n)|1-dimensional nilpotent Lie an-tialgebra with basis {a1, b1, . . . , an, bn; α} that appeared in the case a) of the above proof
is characterized by the relations
ai· bj = δijα, (4.2)
where i, j = 1, . . . , n and all other products of the basic elements vanish. We call it the
Heisenberg antialgebra and denote it by ahn. This algebra is ample.
Remark 4.3. Notice, that the relations (4.2) are exactly as those of the standard Heisen-berg Lie algebra, but the central element α is odd. As in the usual Lie case, the HeisenHeisen-berg antialgebra ahn is a central extension of type II of the abelian Lie antialgebra K2n.
(A2) Another interesting example is a family of Lie antialgebras of dimension 1|2. The basis of these Lie antialgebras will be denoted by {α; a, b}; the commutation relations are
a · b = α, α · a = κ b (4.3)
where κ is a constant and the other products vanish. If κ = 0, then this is just the Heisenberg antialgebra ah1, if κ 6= 0, then this is a non-trivial deformation of ah1.
If K = R, however, the sign of κ is an invariant. We will assume: κ = 1, if K = C, κ = ±1, if K = R. One thus gets two different Lie antialgebras: fah+1 and fah−1.
We are ready to formulate a partial result.
Proposition 4.4. The complete list of the real Lie antialgebras of rank 1 with a
non-degenerate 2-form ω is as follows:
K3, ahn, fah +
1, fah
−
1; (4.4)
in the complex case, the Lie antialgebras fah+1 and fah−1 are isomorphic.
Proof. Consider first the case where a1 is not nilpotent, i.e., α · α 6= 0. We already proved
that, in this case, a is of dimension 1|2, see Lemma 4.2. Therefore, a = K3.
Assume that α · α = 0. If ω is of rank n > 1, one proves, in the same way as in Lemma 4.2, that α · a = 0 for all a, so that a = ahn.
If, finally, ω is of rank 1, then the identity (1.3) implies that (ad1α)2 = 0. One then
easily shows that any such operator is equivalent to ad1α in (4.3) up to the area preserving
changes of the basis. It follows that a = fah+1 or fah−1.
B. The 2-form ω is identically zero The Lie antialgebra a is then non-ample and determined by the operator adα.
(B1) If a1 contains the unit element ε, then a is split into a direct sum (3.19). The
centerless Lie antialgebra a has the basis {ε; a1, . . . , an} with the following set of relations:
ε · ε = ε, ε · ai = 12ai, ai· aj = 0.
We call this Lie antialgebra the affine antialgebra and denote by aaf(n). One then has a= aaf(n) ⊕ Z, where Z is the center of a.
(B2) If α · α = 0 for all α ∈ a0, then ad1α ◦ ad1α = 0. These are very degenerated
Lie antialgebras and their classification is equivalent to the classification of nilpotent (of order 2) linear operators. We do not discuss here this problem of linear algebra.
Let us summarize the above considerations.
Proposition 4.5. A Lie antialgebras of rank 1 with ω = 0 is one of the following two
classes:
a= aaf(n) ⊕ Z, ais of type (B2). (4.5)
C. The “mixed case” 2 < rk ω < dim a0 (C1) Define a 1|3-dimensional Lie
antialge-bra with basis {α; a, b, z} and the relations
and all other products vanish. We denote this Lie antialgebra cah1. The element z spans
the center, so that this is a central extension (of type I) of ah1.
(C2) Define a 1|4-dimensional Lie antialgebra with the basis {α; a, b, z1, z2} and the
relations
a · b = α, α · a = z1, α · b = z2 (4.7)
and all other products vanish. We denote this Lie antialgebra ccah1. This is a central
extension (of type I) of the above algebra cah1.
We are now ready to formulate the main statement of this section.
Theorem 6. A Lie antialgebra of rank 1 is of the form a = a ⊕ Z, where a belongs either
to the list (4.4), or to the list (4.5), or one of the antialgebras cah1, ccah1.
Proof. We already proved the theorem in the following two cases: the form ω is
non-degenerate, or ω ≡ 0. It remains to consider the intermediate case where the 2-form ω is not identically zero but with a non-trivial kernel: I = ker ω 6= {0}. The space I is then an abelian ideal (see Proposition 3.19) and, furthermore, belongs to the center (see Proposition 3.20). We summarize this in a form of a
Lemma 4.6. The Lie antialgebra a is a central extension of a/I.
To complete the classification, one now has to classify the central extensions of type I of the antialgebras with non-degenerate form ω, that is, of (4.2) and (4.3).
Lemma 4.7. The Lie antialgebras ahn with n ≥ 2 and fah1 (resp. fah + 1, fah
−
1) have no
non-trivial central extensions of type I.
Proof. Let C be a 2-cocycle of type I on ahn with n ≥ 2. One has
C (α, ai) = C (aj · bj, ai) = −C (bj · ai, aj) − C (ai· aj, bj) = 0
(from the second identity (3.13)) for all i 6= j. Similarly, C(α, bi) = 0 for all i. Therefore,
C is identically zero.
For the Lie antialgebra fah1 and an arbitrary 2-cocycle C of type I, one has
C (α, b) = κ1C (α, α · a) = 2κ1 C (α · α, a) = 0.
Proof. Indeed, let C be a 2-cocycle of type I on ah1. It is given by the formula
C(α, a) = c1z, C(α, b) = c2z,
where z is an element of the center and c1 and c2 are arbitrary constants. If c1 6= 0, then
choose another element of the basis: b′ = b + c2
c1a. One obtains C(α, b) = 0. Now taking
z′ = c
1z, one gets precisely the Lie antialgebra cah1. This Lie antialgebra is not isomorphic
to ah1 so that the extension is, indeed, non-trivial.
Lemma 4.9. The algebra ccah1 is the unique non-trivial central extension of type I of cah1.
Proof. It is similar to the proof of Lemma 4.8.
In the same way, one proves that the Lie antialgebra ccah1 has no non-trivial central
extensions of type I. We thus classified all the non-trivial central extensions of the Lie antialgebras of rank 1 with non-degenerate form ω.
Theorem 6 is proved.
Acknowledgments. I am pleased to thank K. Bering, F. Chapoton, C. Conley, C. Duval, A. Elduque, Y. Fregier, D. Fuchs, H. Gargoubi, J. Germoni, K. Iohara, O. Kravchenko, M. Kreusch, P. Lecomte, S. Leidwanger, D. Leites, S. Morier-Genoud, S. Parmentier, N. Poncin, C. Roger, S. Tabachnikov, A. Tchoudjem, A. Vaintrob and T. Voronov for enlightening discussions.
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